Body in a bag -Healthcare Management MBA
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A body is discovered in a bag by the river. Authorities initiate an investigation. Using statistics, predictions are made about the offender's profile including that there is a 31.3% probability the offender is a white male, and a 90% probability the offender is white or male. Additional analysis finds a 78.9% probability the victim and offender are the same age group, and a 21% probability they are different age groups. The report concludes with a prediction that the victim is a white male over 18 years old, and the offender is a black male over 18. However, a preliminary forensic report reveals the victim is actually a white female in her 30s. This changes the probability calculations about the offender's profile.
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Body in a bag -Healthcare Management MBA
- 1. T H E B O D Y I N T H E B A G B Y D R S H A LYA S A I N I H C M / 8 7 7 8 4 / 1 5 P R O B A B I L I T Y P R E S E N TA T I O N
- 2. O N E L AT E S P R I N G M O R N I N G ´ ´ ? A stuffed,well worn and heavily stained bag leaning over the river is discovered by elderly Mr Donkins. ? Unidentifiable human body(victim) is discovered inside ? Authorities initiate investigations
- 3. O F F E N D E R P R O F I L E P R E D I C T I O N ? ANS 1-AGE- AT LEAST 18 YEARS-P(A)=6089/7024=0.8668 OR 86.7% ? ANS 2-RACE-WHITE-P(w)=613/1736=0.3531 OR 35.3% ? ANS 3-SEX-MALE-P(m)=6220/7024=0.8855 OR 88.6% ? ANS 4-PROBABILITY OF WHITE+MALE OFFENDER- (USING MULTIPLICATION RULE) P(w)*P(m)=613/1736*6220/7024=0.3531*0.8855=0.3126 OR 31.3% ? ANS 5-PROBABILITY OF WHITE OR MALE OFFENDER-(USING ADDITION RULE)
- 4. O F F E N D E R - W H I T E O R M A L E ? ANS 5 CONTD- ? P(O)=P(W)+P(M)-P(W^M) ? =(613/1736)+(6220?7024)-(6220/7024*613/1736) ? =0.35+0.88-0.33 ? =0.9 OR 90%
- 5. P R O B A B I L I T Y T H AT V I C T I M A N D O F F E N D E R A R E F R O M S A M E O R D I F F E R E N T A G E G R O U P S A G E O F V I C T I M A G E O F O F F E N D E R L E S S T H A N 1 8 AT L E A S T 1 8 U N K N O W N T O TA L L E S S T H A N 1 8 1 1 1 5 8 9 2 8 7 2 8 AT L E A S T 1 8 2 3 1 5 4 2 4 5 4 4 6 1 9 9 U N K N O W N 4 7 6 1 7 9 7 T O TA L 3 4 6 6 0 8 9 5 8 9 7 0 2 4
- 6. A N S 6 - ? USING THE PINK HIGHLIGHTED FIGURES FROM PREVIOUS SLIDE,IN ADDITION THEOREM WE CAN DEDUCE- ? P(O>18 n >18)+P(O>18 n V>18)+p(U>18 n U<18) ? (111/7024)+(5424/7024)+(17/7024) ? =0.015+0.772+0.002 ? =0.789 ? OR 78.9 OR 79 %
- 7. A N S 7 - ? USING THE GREEN HIGHLIGHTED FIGURES FROM THE PREVIOUS SLIDE(SLIDE NO 9),WE CAN DEDUCE USING MULTIPLICATION THEOREM THE PROBABILITY THAT VICTIM AND OFFENDER BELONG TO DIFFERENT AGE GROUPS ? 235/7024*665/7024*572/7024 ? =1472/7024 ? =0.2095 ? 0R 20.95% OR 21%
- 8. ? ALTERNATIVELY,WE CAN ALSO DEDUCE PROBABILITY THAT V AND O BELONG TO DIFFERENT AGE GROUPS BY ? (1-PROBABILITY THAT V AND O BELONG TO SAME AGE GROUP ? (1-0.789) ? =0.211 OR 21%
- 9. J O I N T P R O F I L E P R E D I C T I O N - S A M E A G E C AT E G O RY V I C T I M O F F E N D E R % < 1 8 7 2 8 / 7 0 2 4 1 0 . 3 6 > 1 8 6 1 9 9 / 7 0 2 4 8 8 . 2 5 ? 9 7 / 7 0 2 4 1 . 3 8 % < 1 8 3 4 6 / 7 0 2 4 4 . 9 2 > 1 8 6 0 8 9 / 7 0 2 4 8 6 . 6 8 ? 5 8 9 / 7 0 2 4 8 . 4 ANS 6-LIKELIHOOD THAT VICTIM AND OFFENDER ARE FROM THE SAME AGE GROUP IS CALCULATED IN NEXT SLIDE:
- 10. < 1 8 0 . 1 0 3 6 * 0 . 0 4 9 2 0 . 0 0 5 0 9 0 . 5 % > 1 8 0 . 8 8 2 5 * 0 . 8 6 6 8 8 0 . 7 6 5 0 2 1 6 7 6 . 5 % ? ( U N K N O W N ) 0 . 0 1 3 8 * 0 . 0 8 4 0 . 0 0 1 1 5 0 . 1 1 % (ANS 6 CONTD.)
- 11. R A C E ! R A C E O F V I C T I M R A C E O F T H E O F F E N D E R W H I T E B L A C K O T H E R U N K N O W N T O TA L W H I T E 3 0 7 5 0 1 4 4 4 1 8 9 3 B L A C K 2 2 6 2 8 6 8 4 9 5 6 9 O T H E R 4 7 2 6 1 2 2 4 1 9 9 U N K N O W N 3 3 2 1 2 1 9 7 5 T O TA L 6 1 3 8 3 4 1 7 6 1 1 3 1 7 3 6
- 12. A N S 8 - P R O B A B I L I T Y T H AT V I C T I M A N D O F F E N D E R A R E F R O M S A M E R A C E ( U S I N G A D D I T I O N T H E O R E M ) ? USING THE VALUES HIGHLIGHTED IN PINK IN THE PREVIOUS SLIDE IN ADDITION THEOREM,WE CAN DEDUCE ? P(Ow n Vw)+P(Ob n Vb)+P(Os n Vs)+P(Ou n Vu) ? (307/1736)+(286/1736)+(122/1736)+(19/1736) ? =0.177+0.165+0.07+0.011 ? =0.423 ? OR 42.3%
- 13. A N S 9 - P R O B A B I L I T Y T H AT V I C T I M A N D O F F E N D E R B E L O N G T O D I F F E R E N T R A C E S ? PROBABILITY THAT THE VICTIM AND OFFENDER BELONG TO DIFFERENT RACES AS HIGHLIGHTED IN GREEN IN PREVIOUS SLIDE(SLIDE 12)CAN BE DEDUCED BY USING MULTIPLICATION THEOREM ? (226+47+33)/1736 *(501+26+21)/1736*(44+8+2)/1736*(41+49+4)/1736 ? =306/1736*548/1736*54/1736*94/1736 ? =1002/1736 ? =0.577 ? 0R 57.7%
- 14. ? ALTERNATIVELY,WE CAN DEDUCE THE SAME(V AND O FROM DIFFERENT RACES)BY=1-PROBABILITY THAT V AND O FROM SAME RACE ? =1-0.423 ? =0.577 ? OR 57.7%
- 15. S E X S E X O F V I C T I M S E X O F O F F E N D E R M A L E F E M A L E U N K N O W N T O TA L M A L E 4 4 1 7 4 9 9 7 1 4 9 8 7 F E M A L E 1 7 5 4 1 8 5 2 3 1 9 6 2 U N K O W N 4 9 7 1 9 7 5 T O TA L 6 2 2 0 6 9 1 1 1 3 7 0 2 4
- 16. A N S 1 0 - P R O B A B I L I T Y T H AT V I C T I M A N D O F F E N D E R B E L O N G T O S A M E S E X ? USING ADDING THEOREM WE CAN DEDUCE THE PROBABILITY OF VICTIM AND OFFENDER BELONGING TO THE SAME SEX ? P(Om n Vm)+P(Of n Vf) +P(Ou+Vu) ? (4417/7024)+(185+7024)+(19/7024) ? =0.63+0.03+0.002 ? =0.662 ? OR 66.2%
- 17. A N S 1 1 - P R O B A B I L I T Y T H AT V I C T I M A N D O F F E N D E R B E L O N G T O D I F F E R E N T S E X E S ? USING THE VALUES HIGHLIGHTED IN THE PREVIOUS SLIDE(SLIDE 15)IN GREEN,WE CAN DEDUCE USING MULTIPLICATION THEOREM THE PROBABILITY THAT V AND O BELONG TO DIFFERENT SEXES ? Pc=(1-P) ? (1754+49)/7024*(499+7)/7024*(71+23)/7024 ? =1803+506+94/7024 ? =2403/7024 ? =0.34 ? OR 34%
- 18. ? ALTERNATIVELY,WE CAN DEDUCE PROBABILITY THAT V AND O BELONG TO DIFFERENT SEXES BY=1-PROBABILITY THAT V AND O BELONG TO SAME SEX ? =1-0.662 ? =0.34
- 19. A N S 1 2 - P R O F I L E P R E D I C T I O N - V I C T I M ? ANS 12-AS PER THE GIVEN STATISTICS- ? SEX OF THE VICTIM ? MALE =4987/7024=0.71=71% ? FEMALE=1962/7024=0.28=28% ? UNKNOWN=75/7027=0.01=1% ? AGE OF THE VICTIM ? <18=728/7024=0.1036=10.36% ? >18=6199/7024=0.8825=88.25% ? UNKNOWN=97/7024=0.0138=01.38% ? RACE OF THE VICTIM ? WHITE=893/1736=0.5144=51.44% ? BLACK=569/1736=0.3277=32.77% ? OTHER=199/1736=0.1146=11.46% ? UNKNOWN=75/1736=0.0432=4.32% ? GIVENTHE ABOVE INFO PROFILE PREDICTION FOR VICTIM IS-WHITE,MALE WITH AGE ABOVE 18 YEARS
- 20. A N S 1 3 - P R O F I L E P R E D I C T I O N - O F F E N D E R ? ANS 13-AS PER THE GIVEN STATISTICS- ? SEX OF THE OFFENDER ? MALE =6220/7024=0.8855=88.55% ? FEMALE=691/7024=0.098=9.8% ? UNKNOWN=113/7024=0.016=1.6% ? AGE OF THE OFFENDER ? <18=346/7024=0.04925=4.92% ? >18=6089/7024=0.86688=86.68% ? UNKOWN=589/7024=0.08385=8.4% ? RACE OF THE OFFENDER ? WHITE=613/1736=0.3531=35.31 ? BLACK=834/1736=0.4804=48% ? OTHER=176/1736=0.1013=10.13% ? UNKNOWN=113/1736=0.0651=6.5% ? GIVEN THE ABOVE INFO PROFILE PREDICTION FOR OFFENDER IS-BLACK MALE ,OVER THE AGE OF 18 YEARS
- 21. P R E L I M I N A RY F O R E N S I C R E P O R T F I N D I N G S -No Identification -Death by blunt force trauma -Autopsy revealed body was in water for >2 weeks -White female with blonde hair -Age of the victim=in mid 30s -No signs of child birth -No jewelry recovered
- 22. I M PA C T O F P R E L U M I N A RY F O R E N S I C A N A LY S I S R E P O R T ? Victim profile probability didn't match that of the Forensic Analysis ? Forensic Analysis report-White, female ,>18 years of age(in mid 30s) ? Probability calculation done was-White ,Male,>18 years of age
- 23. A N S 1 4 ? BASED ON THE GIVEN FORENSIC REPORT,THE LIKELIHOOD THAT THE OFFENDER IS ATLEAST 18 ? P(O>18/V>18)=P(O>18 n V>18)/P(V>18) ? (5424/7024)/(6199/7024)=0.875 OR 87.5%
- 24. A N S 1 5 ? LIKELIHOOD THAT THE OFFENDER IS WHITE,BASED ON THE RACE OF THE VICTIM ,USING THE RACE OF VICTIM AND OFFENDER STATISTICS IS- ? P(Ow/Vf)+P(Ow n Vf)/P(Vf) ! ? (1127/7024)/(1962?7024)=0.574 or 57.4% ? HENCE THE LIKELIHOOD OF OFFENDER BEING WHITE IF THE VICTIM IS WHITE(AS PROVED IN FORENSIC REPORT),IS 57.4% %
- 25. R A C E - V I C T I M S E X - O F F E N D E R R E L AT I O N S H I P R A C E O F V I C T I M S E X O F O F F E N D E R M A L E F E M A L E U N K N O W N T O TA L W H I T E 3 1 9 9 3 6 3 4 1 3 6 0 3 B L A C K 2 7 9 3 3 0 5 4 9 3 1 4 7 O T H E R 1 7 9 1 6 4 1 9 9 U N K O W N 4 9 7 1 9 7 5 T O TA L 6 2 2 0 6 9 1 1 1 3 7 0 2 4
- 26. A N S 1 6 ? LIKELIHOOD THAT THE OFFENDER IS MALE,BASED ON THE RACE OF THE VICTIM ,USING THE SEX OF OFFENDER AND RACE OF VICTIM RELATIONSHIP SHOWN IN THE PREVIOUS SLIDE IS ? P(Om/VW)=P(Om n Vw)/P(Vw) ? (3199/7024)/(3603/7024)=0.8878 OR 88.78% ? HENCE IF THE VICTIM IS A FEMALE (AS PROVED IN FORENSIC REPORT)IS 88.78%LIKELY THAT OFFENDER IS A MALE
- 27. A N S 1 7 - ? AS THE FORENSIC REPORT CONFIRMED THAT AGE OF VICTIM IS IN THE SAME GROUP AS AGE PREDICTED VIA PROBABILITY CALCULATION THIS PROBABILITY REMAIN THE SAME ie- OFFENDER over (>)18 YEARS ,GIVEN THAT VICTIM IS OVER (>)18 YEARS
- 28. A N S 1 8 ? GIVEN THAT THE FORENSIC REPORT CONFIRMED THAT VICTIM FALLS IN THE CATEGORY OF AGE DEDUCTED USING PROBABILITY CALCULATION IS CORRECT,THE PROBABILTY THAT VICTIM AND OFFENDER BELONG TO DIFFERENT AGE GROUPS DOESNT CHANGE FROM THE PREVIOUSLY CALCULATED FIGURE
- 29. A N S 1 9 A N D A N S 2 0 ? GIVEN THAT THE RACE OF THE VICTIM DEDUCED IN THE PROBABILTY CALCULATED BASED ON FORENSIC STATISTICS WAS THE SAME AS FOUND IN THE FORENSIC REPORT,THE PROBABILITY THAT ? ANS 19-VICTIM AND OFFENDER BELONG TO SAME RACE REMAINS THE SAME ? ANS 20-PROBABILTYT THAT VICTIM AN DOFFENDER BELONG TO DIFFERENT RACES REMAINS THE SAME AS WELL(AS COMPUTED PRIOR TO EMERGENCE OF THE FORENSIC REPORT)
- 30. A N S 2 1 ? AS THE FORENSIC MEDICAL REPORT DIFFERS FROM THE PROBABILITY OF LIKELY RACE OF THE VICTIM,THE PROBABILITY THAT VICTIM AND OFFENDER BELONG TO SAME SEX ALSO CHANGES BASED ON THE STATISTIC FIGURES ? 185/1962=0.094=9.4%
- 31. A N S 2 2 ? GIVEN THE NEW INFORMATION ON SEX OF THE VICTIM,THE PROBABILITY THAT VICTIM AND OFFENDER BELONG TO DIFFERENT SEXES CAN BE DEDUCED AS FOLLOWS= ? (1-0.094)=0.906 PR 90.6 % ? HENCE ITS VERY LIKELY THAT VICTIM BEING A FEMALE,THE OFFENDER WAS A MALE.
- 32. S E X - V I C T I M R A C E - O F F E N D E R R E L AT I O N S H I P S E X O F V I C T I M R A C E O F O F F E N D E R W H I T E B L A C K O T H E R U N K N O W N T O TA L M A L E 2 1 6 3 2 6 4 2 1 1 1 7 1 4 9 8 7 F E M A L E 1 1 2 7 7 4 9 6 3 2 3 1 9 6 2 U N K N O W N 3 3 2 1 2 1 9 7 5 T O TA L 3 3 2 3 3 4 1 2 1 7 6 1 1 3 7 0 2 4
- 33. A N S 2 3 - N E W P R O F I L E P R E D I C T I O N O F T H E O F F E N D E R ? GIVEN THE FORENSIC REPORT WHICH CONFIRMED THE VICTIM¨S AGE GROUP AND RACE LIE WITHIN THE SAME RANGE AS WAS DEDUCED IN THE PROBABILTY CALCULATION HOWEVER THE SEX OF THE VICTIM WAS FOUND TO BE FEMALE ON MEDICAL EXAMNINTION,USING THE TABLE ATTACHED IN PREVIOUS SLIDE,WE CAN INFER THAT ? AS WITH THE STIPULATED SEX OF VICTIM AS MALE(PROBABILITY CALCULATION DONE PRIOR TO FORENSIC REPORT) THE LIKELIHOOD OF THE OFFENDER BEING MALE WAS THE HIGHEST. ? HOWEVER THE HIGHEST LIKELY RACE FOR OFFENDER ,NOW THAT THE VICTIM IS A CONFIRMED FEMALE,CHANGES TO WHITE.
- 34. A N S 2 4 B A S E D O N T H E G I V E N S TAT I S T I C S ! 4 5 % - O F F E N D E R S A R E R E L AT E D T O O R A C Q U A I N T E D T O T H E V I C T I M 1 5 % - O F F E N D E R S A R E U N R E L AT E D T O T H E V I C T I M 4 0 % - O F F E N D E R S H AV E U N K N O W N R E L AT I O N S H I P S T O T H E V I C T I M W E C A N P R E D I C T I T I S P R O B A B L E T H AT T H E O F F E N D E R WA S K N O W N T O T H E V I C T I M .